Search any question & find its solution
Question:
Answered & Verified by Expert
If $\vec{a}=t \vec{b}$ where $t < 0$ is a scalar, then
Options:
Solution:
1836 Upvotes
Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ are unlike vectors and either $|\overrightarrow{\mathrm{a}}| \geq|\overrightarrow{\mathrm{b}}|$ or $|\overrightarrow{\mathrm{a}}| < |\overrightarrow{\mathrm{b}}|$
Given $\vec{a}=t \vec{b}$ where $t < 0$ is a scalar
$\therefore \mathrm{t}$ is negative.
$\Rightarrow \mathrm{t}=-\mathrm{k}$ for some $\mathrm{k} \in \mathrm{Z}^{+}$
So, we have $\vec{a}=-k \vec{b}$ ......(i)
$\therefore \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ are unlike vectors.
Taking mod on both sides of eqn. (i), we get
$$
|\vec{a}|=|-k \vec{b}| \Rightarrow|\vec{a}|=|k||\vec{b}|
$$
If $0 < |k| < 1 \mid$ then $|\vec{a}| < |\vec{b}|$
And if $|\mathrm{k}| \geq 1$ then $|\vec{a}| \geq|\vec{b}|$
$\therefore \mathrm{t}$ is negative.
$\Rightarrow \mathrm{t}=-\mathrm{k}$ for some $\mathrm{k} \in \mathrm{Z}^{+}$
So, we have $\vec{a}=-k \vec{b}$ ......(i)
$\therefore \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ are unlike vectors.
Taking mod on both sides of eqn. (i), we get
$$
|\vec{a}|=|-k \vec{b}| \Rightarrow|\vec{a}|=|k||\vec{b}|
$$
If $0 < |k| < 1 \mid$ then $|\vec{a}| < |\vec{b}|$
And if $|\mathrm{k}| \geq 1$ then $|\vec{a}| \geq|\vec{b}|$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.